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5. Solve the inequality $\arcsin \left(\frac{5}{2 \pi} \arccos x\right)>\arccos \left(\frac{10}{3 \pi} \arcsin x\right)$. | Answer: $\left[\cos \frac{2 \pi}{5} ; \cos \frac{8 \pi}{25}\right) \cup\left(\cos \frac{8 \pi}{25} ; \cos \frac{\pi}{5}\right]$.
Solution. Let $u=\frac{5}{2 \pi} \arccos x, v=\frac{10}{3 \pi} \arcsin x$. Then the relation $4 u+3 v=5$ holds. The solution to the inequality $\arcsin u>\arccos v$ is the set
$$
0 \leqslan... | [\cos\frac{2\pi}{5};\cos\frac{8\pi}{25})\cup(\cos\frac{8\pi}{25};\cos\frac{\pi}{5}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,090 |
6. Find all sets of numbers $x_{1}, x_{2}, \ldots, x_{n+1}$ such that $x_{1}=x_{n+1}$ and for all $k=1, \ldots, n$ the equality
$$
2 \log _{2} x_{k} \cdot \log _{2} x_{k+1}-\log _{2}^{2} x_{k}=9
$$ | Answer: $x_{k}=8, k=1, \ldots, n+1$, or $x_{k}=\frac{1}{8}, k=1, \ldots, n+1$.
Solution. We need to solve the system of equations
$$
\left\{\begin{array}{l}
9-2 \log _{2} x_{1} \cdot \log _{2} x_{2}+\log _{2}^{2} x_{1}=0 \\
9-2 \log _{2} x_{2} \cdot \log _{2} x_{3}+\log _{2}^{2} x_{2}=0 \\
\cdots \\
9-2 \log _{2} x_{... | x_{k}=8,k=1,\ldots,n+1,orx_{k}=\frac{1}{8},k=1,\ldots,n+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,091 |
7. Find the maximum value of the function
$$
f(x)=\sin (x+\sin x)+\sin (x-\sin x)+\left(\frac{\pi}{2}-2\right) \sin \sin x
$$ | Answer: $\frac{\pi-2}{\sqrt{2}}$.
Solution. Using the formula for converting the sum of sines into a product, we get
$$
f(x)=2 \sin x \cdot \cos \sin x+\left(\frac{\pi}{2}-2\right) \sin \sin x=g(\sin x)
$$
where $g(t)=2 t \cos t+\left(\frac{\pi}{2}-2\right) \sin t, t=\sin x \in[-1 ; 1]$. The function $g$ is odd and
... | \frac{\pi-2}{\sqrt{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 6,092 |
8. Andrei likes all numbers that are not divisible by 3, and Tanya likes all numbers that do not contain digits divisible by 3.
a) How many four-digit numbers are liked by both Andrei and Tanya?
b) Find the total sum of the digits of all such four-digit numbers. | Answer: a) 810; b) 14580.
Solution. a) The required numbers must be composed of the digits $1,2,4,5,7,8$, and according to the divisibility rule by 3, the sum of the digits in each number should not be divisible by three. The digits 1, 4, and 7 (let's call them the digits of set $A$) give a remainder of 1 when divided... | 810 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,093 |
1. In a certain country, the alphabet consists of three letters: "M", "G", and "U". A word is any finite sequence of these letters, in which two consonants cannot stand next to each other and two vowels cannot stand next to each other. How many 200-letter words are there in this country that contain each of the three l... | Answer: $2^{101}-4$. Solution. After understanding the structure of the words, we consider three cases.
1) The first letter is M. Then the second is U, the third is G or M, the fourth is U, the fifth is G or M, ..., the 200th is U. There will be $2^{99}$ such words. 2) Similarly, if the first letter is G.
2) The first... | 2^{101}-4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,094 |
2. The graph of the quadratic function $f(x)=x^{2}+2 p x-p^{2}+7 p-2015$ intersects the coordinate axes at three points: $A, B$ and $C$. Find the value of $p$ for which the product of the lengths of the segments $O A \times O B \times O C$ is minimized. | Answer: $p=\frac{7}{2}$. Solution. For the parabola to intersect at three points, it is necessary for the equation to have two distinct roots $x_{1}$ and $x_{2}$. It is not difficult to verify that the discriminant of the given polynomial is always positive. Then the product $O A \times O B \times O C=\left|x_{1}\right... | \frac{7}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,095 |
3. In triangle $\triangle A B C$, the base $A B$ of which lies on the x-axis, altitudes $A M, B N$ and $C K$ are drawn. Find the length of the base $A B$, if the coordinates of points $M(2,2)$ and $N(4,4)$ are known. | Answer: $A B=4 \sqrt{5}$. Solution.
The circle that is based on $A B$ as its diameter contains points $M$ and $N$ (see the figure). Its center $D$ is equidistant from $M$ and $N$. Since the ... | 4\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,096 |
4. All natural numbers are divided into "good" and "bad" according to the following rules:
a) From any bad number, you can subtract some natural number not exceeding its half so that the resulting difference becomes "good".
b) From a "good" number, you cannot subtract no more than half of it so that it remains "good"... | Answer: 2047. Solution. Considering the first few natural numbers, we notice that good numbers have the form $2^{n}-1$ (while $2^{n}, \ldots, 2 n+1-2$ are bad).
We will prove this by mathematical induction. For $n=1$, the statement is given in the condition. Suppose the statement is proven for $n-1$. Consider a number... | 2047 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,097 |
5. In a convex quadrilateral $A B C D$, the areas of triangles $A B D$ and $B C D$ are equal, and the area of $A C D$ is half the area of $A B D$. Find the length of the segment $C M$, where $M$ is the midpoint of side $A B$, if it is known that $A D=12$. | Answer: $C M=18$. Solution. From the equality of the areas of $A B C$ and $C B D$, it follows that $O$ - the intersection point of the diagonals $A C$ and $B D$ bisects $A C$ (see figure). And from the fact that $S(A C D) = \frac{1}{2} S(A B D)$, it follows that $S(A O D) = \frac{1}{3} S(A O B)$, therefore, $O D = \fra... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,098 |
6. The sequence is defined by the recurrence relation: $a_{1}=a_{2}=1$, $a_{n+1}=a_{n} \cdot a_{n-1}+3$. Can the number $a_{2015}-a_{2011}-39$ be prime? | Answer: No, this number will be divisible by $a_{2011}>1$. Solution. We will show that this number will be divisible by $a_{2011}$. Indeed, consider the remainders of the subsequent terms when divided by $a_{2011}$. Obviously, $a_{2012}$ and $a_{2013}$ give remainders $3, a_{2014}-12$, and $a_{2015}-39$. Therefore, $a_... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,099 |
1. Find the smallest natural number that is greater than the sum of its digits by 1755 (the year of the founding of Moscow University). | Answer: 1770.
Solution. From the condition, it follows that the desired number cannot consist of three or fewer digits. We will look for the smallest such number in the form $\overline{a b c d}$, where $a, b, c, d$ are digits, and $a \neq 0$. We will form the equation:
$$
\begin{gathered}
\overline{a b c d}=a+b+c+d+1... | 1770 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,100 |
2. Two pieces of cheese have the shape of a rectangular parallelepiped each. The length of the first piece is $50\%$ greater than the length of the second piece, while the width and height of the first piece are $20\%$ and $30\%$ less than the width and height of the second piece. Which piece of cheese has a larger vol... | Answer: The second one is $19 \frac{1}{21} \%$ more (the first one is $16 \%$ less).
Solution. Let $a, b$ and $c$ be the length, width, and height of the second piece of cheese, respectively. Then its volume is $V_{2}=a b c$. According to the problem, the volume of the first piece of cheese is $V_{1}=\frac{3}{2} a \cd... | 19\frac{1}{21} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,101 |
3. Having walked $2 / 5$ of the length of a narrow bridge, a pedestrian noticed that a car was approaching the bridge from behind. Then he walked back and met the car at the beginning of the bridge. If the pedestrian had continued walking forward, the car would have caught up with him at the end of the bridge. Find the... | Answer: 5.
Solution. In the time $t$ that the pedestrian walked towards the car until they met at the beginning of the bridge, he covered $2 / 5$ of the bridge's length. Therefore, if the pedestrian continued walking forward, in time $t$ he would have covered another $2 / 5$ of the bridge's length, and he would have $... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,102 |
5. The height of a right triangle, dropped to its hypotenuse, divides the bisector of an acute angle in the ratio 5 : 2, counting from the vertex. Find the measure of this angle.
Answer: $\arccos \frac{5}{7}$. | Solution. Let $A D=5 x, D L=2 x$. Then $\cos \angle L A H=$ $=\frac{A H}{A D}=\frac{A H}{5 x}=\cos \angle L A C=\frac{A C}{A L}=\frac{A C}{7 x}$. Therefore,
$\cos \angle C A H=\frac{A H}{A C}... | \arccos\frac{5}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,103 |
6. How many solutions does the equation
$$
\frac{1}{(x-1)^{2}}+\frac{1}{(x-2)^{2}}=\frac{2}{x^{2}} ?
$$ | Answer: 1.
Solution. Let us consider the functions
$$
f(x)=\frac{1}{(x-1)^{2}}+\frac{1}{(x-2)^{2}} \quad \text { and } \quad g(x)=\frac{2}{x^{2}}.
$$
For $x < 0$, both functions $f$ and $g... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,104 |
7. Given three points, the distances between which are 4, 6, and 7. How many pairwise distinct triangles exist for which each of these points is either a vertex or the midpoint of a side? | Answer: 11.
Solution. Let's list all the constructions of triangles that satisfy the condition of the problem, indicating the lengths of the sides.
| | Description of the Triangle | | Lengths of Sides |
| :---: | :---: | :---: | :---: |
| №1 | All points are vertices | | $4,6,7$ |
| | One point is a vertex, two a... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,105 |
8. In a commercial football tournament, five teams participated. Each was supposed to play exactly one match against each other. Due to financial difficulties, the organizers canceled some games. In the end, it turned out that all teams had scored a different number of points, and no team had a zero in the points score... | Answer: 6.
Solution. The minimum possible total score: $1+2+3+4+5=15$. In one game, a maximum of 3 points (in total) can be scored. Therefore, there were at least 5 games. However, if there were exactly 5 games, then all games would have to end with one of the teams winning, and then no team would have scored exactly ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,106 |
9. Inside a circle, distinct points $A$ and $B$ are given. Where on the circle should point $C$ be located so that the angle $ACB$ is as large as possible? List all options and justify that there are no others. | Answer: one of the points (or both points) of tangency of the circle passing through $A$ and $B$ with the original circle. There are two such circles, and the one for which the angle is larger should be chosen. Solution. Draw the line $c$ through points $A$ and $B$. Construct circles $\omega_{1}$ and $\omega_{2}$, pass... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,107 |
10. What is greater: $\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\ldots+\frac{1}{2010}-\frac{1}{2011}$ or $\frac{1}{1006}+\frac{1}{1007}+\ldots+\frac{1}{2011}$? | Answer: the right number is greater.
Solution. Let's denote the sum of the form $1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$ as $H_{n}$. Then the left part can be transformed as follows:
$$
\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\ldots+\frac{1}{2010}-\frac{1}{2011}=2 \cdot\left(\frac{1}{2}+\frac{1}{4}+\ldo... | \frac{1}{1006}+\frac{1}{1007}+\ldots+\frac{1}{2011}>\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,108 |
1. There were 21 consecutive natural numbers written on the board. When one of the numbers was erased, the sum of the remaining numbers became 2017. Which number was erased? | Answer: 104.
Solution: Let the numbers on the board be N-10, N-9,..,N, ..., N+10. Their sum is $21 \mathrm{~N}$. When one of these numbers - x - was erased, the sum became 2017, $21 \mathrm{~N}-$ $x=2017$. Therefore, $x=21 N-2017$, since this is one of these numbers, we get $N-10 \leq 21 N-2017 \leq N+10$. Solving the... | 104 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,109 |
2. A line passes through the point with coordinates $(10 ; 0)$ and intersects the parabola $y=$ $x^{2}$ at points with abscissas $x_{1}$ and $x_{2}$. Find $\frac{1}{x_{1}}+\frac{1}{x_{2}}$. | Answer: 0.1.
Solution. The points $\mathrm{x}_{1}, \mathrm{x}_{2}$, are solutions to the equation $x^{2}=k(x-10)$, where $\mathrm{k}$ is the slope coefficient of the line. Then, by Vieta's theorem, $x_{1}+x_{2}=k, x_{1} x_{2}=10 k$. Therefore, $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=\frac{k}{1... | 0.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,110 |
3. How many diagonals in a regular 32-sided polygon are not parallel to any of its sides | Answer: 240.
Solution: In a 32-sided polygon, there are $32*(32-3)/2=464$ diagonals in total. We can divide the sides into 16 pairs of parallel sides. It is not hard to notice that if we fix a pair, i.e., 4 vertices, the remaining vertices can be connected in pairs by diagonals parallel to this pair. There will be a t... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,111 |
4. For natural numbers $m$ and $n$, it is known that $3 n^{3}=5 m^{2}$. Find the smallest possible value of $m+n$. | Answer: 60.
Solution: Obviously, if $m, n$ contain prime factors other than 3 or 5, then they can be canceled out (and reduce $m+n$).
Let $n=3^{a} \cdot 5^{b}, m=3^{c} \cdot 5^{d}$. Then the condition implies that $3 a+1=2 c, 3 b=2 d+1$.
The smallest possible values are: $a=1, b=1, c=2, d=1$, from which $n=15, m=45$... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,112 |
5. Pete and Vasya are playing a game. On the board, there is a number: 11223334445555666677777. In one move, it is allowed to erase any number of identical digits. The player who erases the last digit wins. Pete goes first. Can he play in such a way as to guarantee a win? | Answer: Yes, he can.
Solution: On his first move, Petya can, for example, erase all the digits 7. We write the remaining digits as follows:
After this, Petya symmetrically (relative to the middle line) repeats Vasya's moves. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,113 |
7. For the function $y=f(x)$, it is known that it is defined and continuous on the entire number line, odd, and periodic with a period of 5, and that $f(-1)=f(2)=-1$. What is the minimum number of roots that the equation $f(x)=0$ can have on the interval [1755; 2017]? Answer: 210. | Solution. Since the function $f$ is odd and defined at zero, we get $f(0)=-f(0)$ $\Rightarrow f(0)=0$. Due to the 5-periodicity, we then have $f(5)=f(0)=0$. Using the oddness again: $f(1)=-f(-1)=1$, and due to the 5-periodicity $f(3)=f(-2)=1, f(4)=$ $f(-1)=-1$. Thus, at points $1, 2, 3$, and $4$, the values of the func... | 210 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,115 |
1. When a car travels from point $A$ to point $B$, it spends $25 \%$ of the time on the uphill journey, $60 \%$ on the flat road, and the remaining time on the downhill journey. The time it takes to travel from $A$ to $B$ and back from $B$ to $A$ along the same road is the same, and its speeds uphill, downhill, and on ... | Answer: $5 / 3$.
Solution. If the speed of the car downhill is $x \neq 1$ times the speed uphill, then from the condition we have:
$$
\begin{aligned}
0.25+0.15= & 0.25 / x+0.15 x \quad \Rightarrow \quad 0.15 x^{2}-(0.25+0.15) x+0.25=0 \\
& \Rightarrow \quad(x-1)(x-5 / 3)=0 \quad \Rightarrow \quad x=5 / 3
\end{aligned... | \frac{5}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,116 |
2. Solve the equation $\sqrt{\sqrt{x+2}+\sqrt{x-2}}=2 \sqrt{\sqrt{x+2}-\sqrt{x-2}}+\sqrt{2}$.
Answer: $\frac{257}{16}$. | Solution. Let $t=\sqrt{x+2}+\sqrt{x-2}$, then $\sqrt{\sqrt{x+2}-\sqrt{x-2}}=\frac{2}{\sqrt{t}}$, and the given equation takes the form $\sqrt{t}=\frac{4}{\sqrt{t}}+\sqrt{2}$, or $t-\sqrt{2 t}-4=0$. Solving this equation as a quadratic in $\sqrt{t}$, we find $\sqrt{t}=-\sqrt{2}$ or $\sqrt{t}=2 \sqrt{2}$. The first equat... | \frac{257}{16} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,117 |
3. Determine which of the numbers is greater: $11^{\lg 121}$ or $10 \cdot 10^{\lg ^{2}} 11+11$. | Answer: the first.
Solution. Since $11^{\lg 121}=\left(11^{\lg 11}\right)^{2}$, and $10 \cdot 10^{\lg ^{2} 11}+11=10 \cdot\left(10^{\lg 11}\right)^{\lg 11}+11=10 \cdot 11^{\lg 11}+11$, we need to compare the numbers $x^{2}$ and $10 x+11$, where $x=11^{\lg 11}$. The expression $x^{2}-10 x-11=(x-11)(x+1)$ at the point $... | thefirst | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,118 |
4. Quadrilateral $ABCD$ is inscribed in a circle. The tangents to this circle, drawn at points $A$ and $C$, intersect on the line $BD$. Find the side $AD$, if $AB=2$ and $BC: CD=4: 5$.
Answer: $\frac{5}{2}$. | Solution. Let $M$ be the point of intersection of the tangents. By the condition, $M$ lies on the line $B D$. Since $B C < C D$, point $B$ is between points $M$ and $D$. Then, since $\angle M C B = \angle C D B = \frac{\smile B C}{2}$, triangles $M C B$ and $M D C$ are similar, so $\frac{C D}{B C} = \frac{M D}{M C}$. S... | \frac{5}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,119 |
5. Calculate $\sqrt{n}+\sqrt{n+524}$, given that this number is rational and that $n$ is a natural number. | Answer: 262.
Solution. Let the desired number be $a$. We have $\sqrt{n+524}=a-\sqrt{n}, n+524=a^{2}-2 a \sqrt{n}+n$. By the condition, $a$ is rational, so $\sqrt{n}$ is also rational. Therefore, $n=k^{2}, k \in \mathbb{N}$. Then the number $\sqrt{n+524}$ is also rational, so $n+524=m^{2}, m \in \mathbb{N}$. Thus, $m^{... | 262 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,120 |
6. A sphere is inscribed in a right circular cone, the radius of the base of which is equal to 2. Find the volume of this sphere if it is three times smaller than the volume of the cone. | Answer: $\frac{16 \pi}{9} \sqrt{9+5 \sqrt{3}}$ or $\frac{16 \pi}{9} \sqrt{9-5 \sqrt{3}}$.
Solution. Let $r$ be the radius of the inscribed sphere, $R=2$ be the radius of the base, $h$ be the height of the cone, $2 \varphi$ be the angle between the slant height of the cone and the plane of its base, $0<\varphi<\pi / 4,... | \frac{16\pi}{9}\sqrt{9+5\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,121 |
7. Find all values of $a$ for each of which exactly one of the following two statements is true:
1) «The equation $\cos (\cos x)+\sin (\sin x)=a$ has exactly two roots on the interval $[0 ; \pi]$ »
2) «The equation $\sin ^{4} x+\cos ^{4} x+\sin 2 x=a$ has roots».
Answer: $\left[-\frac{1}{2} ; \cos 1\right) \cup\left(\... | Solution. 1) The function $f(x)=\cos (\cos x)+\sin (\sin x)$ increases on the interval from 0 to $\frac{\pi}{2}$ (each of the terms is a monotonically increasing function) and decreases on the interval from $\frac{\pi}{2}$ to $\pi$ (since $f(\pi-x)=f(x)$). Therefore, $E(f)=[f(0) ; f(\pi / 2)]=[\cos 1 ; 1+\sin 1]$. This... | [-\frac{1}{2};\cos1)\cup(\frac{3}{2};1+\sin1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,122 |
8. We consider all possible sets consisting of 2017 different natural numbers, where in each set no number can be represented as the sum of two other numbers in the set. What is the smallest value that the largest number in such a set can take?
Answer: 4032. | Solution. Let for brevity $n=2017$. Consider the following set of $n$ numbers: $n-1, n, n+1, \ldots$, $2 n-3, 2 n-2$. Since $(n-1) + n = 2 n - 1 > 2 n - 2$, no number in this set can equal the sum of two others, meaning the given set satisfies the problem's condition.
Now, let there be an arbitrary set of $n$ natural ... | 4032 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,123 |
1. In a full container, there are 150 watermelons and melons for a total of 24 thousand rubles, with all the watermelons together costing as much as all the melons. How much does one watermelon cost, given that the container can hold 120 melons (without watermelons) and 160 watermelons (without melons)? | Answer: 100 rubles. Solution. Let there be $x$ watermelons, then there will be $150-x$ melons. If a container can hold 120 melons, then one melon occupies $\frac{1}{120}$ of the container. Similarly, one watermelon occupies $\frac{1}{160}$ of the container. Therefore, if the container is fully loaded with 150 watermelo... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,124 |
2. For two positive numbers $a \neq b$, it is known that
$$
a^{2}-2015 a=b^{2}-2015 b
$$
What is the smallest value that $a^{2}+b^{2}$ can take? | Answer: $2015^{2} / 2=2030112.5$. Solution. By moving and reducing by $a-b \neq$ 0, we get $a+b=2015$. Then $a^{2}+b^{2}=a^{2}+(2015-a)^{2}=2 a^{2}-$ $2030 a+2015^{2}$ and the minimum value is achieved at the vertex of the parabola $2 x^{2}+4030 x+2015^{2}$, i.e., when $a=b=2015 / 2$. | 2030112.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,125 |
3. A $3 \times 3$ table needs to be filled with the numbers 2014, 2015, and 2016, such that the sum of the numbers in each row is the same. In how many different ways can this be done? | Answer: 831. Solution. Subtract 2015 from all numbers - the sums will remain the same, and the numbers in the table will take on values of 0 and $\pm 1$. Consider the 27 possible combinations of numbers in one row. Possible values of the sum of these numbers: $\pm 3$ - 1 combination each, $\pm 2$ - 3 combinations each,... | 831 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,126 |
4. At Andrei's birthday, Yana, who arrived last, gave him a ball, and Eduard, who arrived second to last, gave him a calculator. Testing the calculator, Andrei noticed that the product of the total number of his gifts and the number of gifts he had before Eduard arrived is exactly 16 more than the product of his age an... | Answer: 18. Solution. If $n$ is the number of gifts, and $a$ is Andrei's age, then $n(n-2)=a(n-1)$. From this, $a=\frac{n^{2}-2 n-16}{n-1}=n-1-\frac{17}{n-1}$. Therefore, $n-1=17, a=16$. | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,127 |
5. In an equilateral triangle $A B C$, points $A_{1}$ and $A_{2}$ are chosen on side $B C$ such that $B A_{1}=A_{1} A_{2}=A_{2} C$. On side $A C$, a point $B_{1}$ is chosen such that $A B_{1}: B_{1} C=1: 2$. Find the sum of the angles $\angle A A_{1} B_{1}+\angle A A_{2} B_{1}$. | Answer: $30^{\circ}$. Solution. Since $A_{1} B_{1} \| A B$, then $\angle B A A_{1}=\angle A A_{1} B_{1}$. From symmetry, $\angle B A A_{1}=\angle C A A_{2}$. It remains to note that $\angle C B_{1} A_{2}=$ $\angle B_{1} A A_{2}+\angle A A_{2} B_{1}$ as an exterior angle in $\triangle A A_{2} B_{1}$. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,128 |
6. Find the greatest possible value of $\gcd(x+2015 y, y+2015 x)$, given that $x$ and $y$ are coprime numbers. | Answer: $2015^{2}-1=4060224$. Solution. Note that the common divisor will also divide $(x+2015 y)-2015(y+2015 x)=\left(1-2015^{2}\right) x$. Similarly, it divides $\left(1-2015^{2}\right) y$, and since $(x, y)=1$, it divides $\left(1-2015^{2}\right)$. On the other hand, if we take $x=1, y=2015^{2}-2016$, then we get $\... | 4060224 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,129 |
4. A flea jumps along the number line, and the length of each jump cannot be less than $n$. It starts its movement from the origin and wants to visit all integer points belonging to the segment $[0,2013]$ (and only them!) exactly once. For what greatest value of $n$ will it be able to do this? | Answer: $n=1006$.
Solution: For $n=1006$, a path can be constructed as follows:
$$
0 \rightarrow 1007 \rightarrow 1 \rightarrow 1008 \rightarrow \ldots \rightarrow 1005 \rightarrow 2012 \rightarrow 1006 \rightarrow 2013
$$
We will prove that $n$ cannot be greater than 1006. Indeed, suppose $n \geqslant 1007$. Then t... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,130 |
5. Solve the cryptarithm (the same letters correspond to the same digits, different letters correspond to different digits)
$$
\mathrm{MSU}+\mathrm{MSU}+\mathrm{MSU}+\mathrm{MSU}+\mathrm{OLYMP}+\mathrm{OLYMP}=\mathrm{MOSCOW}
$$ | Answer: $C=5, L=7, M=1, O=9, P=2, S=4, U=3, W=6, Y=0,143+143+$ $143+143+97012+97012=194596$.
Solution: Note that since $M S U \leq 987, O L Y M P \leq 98765$, then
$M S U+M S U+M S U+M S U+O L Y M P+O L Y M P \leq 987 \times 4+98765 \times 4=201478$.
Thus, $M$ can only be 1 or 2. $M=2$ does not work, because in this... | 143+143+143+143+97012+97012=194596 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,131 |
6. In how many different ways can a chess king move from square $e 1$ to square $h 5$, if it is allowed to move only one square to the right, up, or diagonally to the right-up? | Answer: 129 ways.
Solution: Sequentially (starting from e1) find the number of ways to reach each cell. Each number (except 1) is obtained by summing the neighbors below, to the left, and diagonally left-below.
^{2}+\left|\log _{2} \frac{120-2 x \sqrt{32-2 x}}{\left(x^{2}-2 x+8\right)^{3}}\right|}{5 \log _{7}(71-2 x \sqrt{32-2 x})-2 \log _{2}(120-2 x \sqrt{32-2 x})} \geqslant 0
$$ | Answer: $-13-\sqrt{57} \approx-20.55$.
Solution. Let $f(x)=-\log _{2}(x+120)$. Then, taking into account the condition $71-2 x \sqrt{32-2 x}>0$, the original inequality can be rewritten as
$$
\frac{2 f(-2 x \sqrt{32-2 x})+\left|3 f\left(x^{2}-2 x-112\right)-f(-2 x \sqrt{32-2 x})\right|}{\left(2 \log _{2} 7\right) \cd... | -13-\sqrt{57} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,133 |
Task 1. Two cars covered the same distance. The speed of the first was constant and three times less than the initial speed of the second. The second car traveled the first half of the distance without changing its speed, then it suddenly reduced its speed by half, traveled another quarter of the distance at a constant... | Answer: $\frac{5}{3}$.
Solution. Let $V_{1}$ be the speed of the first car, $V_{0}$ the initial speed of the second, and the distance covered by each car be $S$. Then $V_{0}=3 V_{1}$. The first car spent the entire journey time $t_{1}=\frac{S}{V_{1}}$, while the second car spent time
$$
t_{2}=\frac{\frac{S}{2}}{V_{0}... | \frac{5}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,134 |
Problem 2. How many divisors of the number $2021^{2021}$ have a cube root that is a natural number? | Answer: 454276.
Solution. Since $2021=43 \cdot 47$, all divisors of the number $2021^{2021}$ have the form $43^{\alpha} \cdot 47^{\beta}$, where $\alpha, \beta \in[0 ; 2021]$. In this case, the exact cubes are numbers of the form $43^{3 n} \cdot 47^{3 k}$, where $3 n, 3 k \in[0 ; 2021]$, that is, $n, k \in[0 ; 673]$. ... | 454276 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,135 |
Problem 3. Solve the system
$$
\left\{\begin{array}{l}
\left|x^{4}-625 x^{2}\right| \neq x^{4}-625 x^{2} \\
\left|6 x^{2}-257 x+251\right|+6 x^{2}-257 x+251=0
\end{array}\right.
$$ | Answer: $[1 ; 25)$.
Solution. The original system is equivalent to the system
$$
\left\{\begin{array}{l}
x^{4}-625 x^{2}<0 \\
6 x^{2}-257 x+251 \leqslant 0
\end{array}\right.
$$
which has the solution: $[1 ; 25)$. | [1;25) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,136 |
Problem 4. How many triples of numbers $a, b, c$ exist, each of which is a root of the corresponding equation $a x^{2}+b x+c=0$? | Answer: 5.
Solution. If $a=0$ or $a=b=c$, then $a=b=c=0$. Otherwise: if $a=b \neq c$, then either $a=b=-1, c=0$, or $a=b=1, c=-2$; if $a \neq b=c$, then either $a=1, b=c=-0.5$; if $a=c \neq b$, then $a=c=c_{0}, b=1 / c_{0}$, where the number $c_{0}<0$ is the unique root of the equation $c^{3}+c=-1$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,137 |
Problem 6. In an acute-angled triangle $A B C$ with side $A C=1$, the altitude $B H$ is drawn, in triangle $B H C$ - the bisector $C L$, in triangle $B L C$ - the median $B M$. The line $A L$ intersects side $B C$ at point $K$, and $\angle B H K=\angle M H C=15^{\circ}$. Find the area of quadrilateral $K L H M$. | Answer: $\frac{3(2-\sqrt{3})}{8}$.
Solution. Triangle $C H L$ is a right triangle, so its vertices lie on a circle with the center at the midpoint of the hypotenuse - point $M$. Then $M H = M C$ and $\angle L H K = \angle M H C = \angle M C H = \angle M C K = 15^{\circ}$. Therefore, point $K$ also lies on the same cir... | \frac{3(2-\sqrt{3})}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,139 |
1.1. (2 points) From the digits 1, 3, and 5, different three-digit numbers are formed, each of which has all distinct digits. Find the sum of all such three-digit numbers. | Answer: 1998.
Solution. All possible numbers: $135,153,315,351,513,531$. | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,140 |
2.1. (2 points) In a triangle, one of the angles is less than $50^{\circ}$, and another is less than $70^{\circ}$. Find the cosine of the third angle if its sine is $\frac{4}{7}$. | Answer: $-\frac{\sqrt{33}}{7} \approx-0.82$.
Solution. If $\sin \alpha=\frac{4}{7}$ and $\alpha>180^{\circ}-\left(50^{\circ}+70^{\circ}\right)=60^{\circ}$, then $|\cos \alpha|=\sqrt{1-\sin ^{2} \alpha}=\frac{\sqrt{33}}{7}$ and $\cos \alpha<\frac{1}{2}$, therefore $\cos \alpha=-\frac{\sqrt{33}}{7}$. | -\frac{\sqrt{33}}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,141 |
2.2. In a triangle, one of the angles is less than $40^{\circ}$, and another is less than $80^{\circ}$. Find the cosine of the third angle if its sine is $\frac{5}{8}$. | Answer: $-\frac{\sqrt{39}}{8} \approx-0.78$. | -\frac{\sqrt{39}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,142 |
3.1. (12 points) The number
$$
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{2017}{2018!}
$$
was written as an irreducible fraction with natural numerator and denominator. Find the last two digits of the numerator. | Answer: 99.
Solution. We have
$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{2017}{2018!}=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\ldots+\left(\frac{1}{2017!}-\frac{1}{2018!}\right)=1-\frac{1}{2018!}=\frac{2018!-1}{2018!}$.
In the end, we obtained an irreducible fraction, and the las... | 99 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,143 |
4.1. (12 points) In rectangle $A B C D$, point $E$ is marked on the extension of side $C D$ beyond point $D$. The bisector of angle $A B C$ intersects side $A D$ at point $K$, and the bisector of angle $A D E$ intersects the extension of side $A B$ at point $M$. Find $B C$, if $M K=8$ and $A B=3$. | Answer: $\sqrt{55} \approx 7.42$.
Solution. Since $B K$ is the bisector of angle $A B C$, triangle $A B K$ is isosceles, $A B = A K$. Similarly, we get that $A D = A M$. Therefore, triangles $A K M$ and $A B D$ are equal by two legs. Applying the Pythagorean theorem in triangle $A K M$, we find $B C = A D = A M = \sqr... | \sqrt{55} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,144 |
4.2. In rectangle $A B C D$, point $E$ is marked on the extension of side $C D$ beyond point $D$. The bisector of angle $A B C$ intersects side $A D$ at point $K$, and the bisector of angle $A D E$ intersects the extension of side $A B$ at point $M$. Find $B C$, if $M K=9$ and $A B=4$. | Answer: $\sqrt{65} \approx 8.06$. | \sqrt{65} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,145 |
4.3. In rectangle $A B C D$, point $E$ is marked on the extension of side $C D$ beyond point $D$. The bisector of angle $A B C$ intersects side $A D$ at point $K$, and the bisector of angle $A D E$ intersects the extension of side $A B$ at point $M$. Find $B C$, if $M K=10$ and $A B=7$. | Answer: $\sqrt{51} \approx 7.14$. | \sqrt{51} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,146 |
5.1. (12 points) The decreasing sequence $a, b, c$ is a geometric progression, and the sequence $19 a, \frac{124 b}{13}, \frac{c}{13}$ is an arithmetic progression. Find the common ratio of the geometric progression. | Answer: 247.
Solution. Let $b=a q, c=a q^{2}$. The properties of an arithmetic progression and the conditions of the problem lead to the equation $2 \cdot \frac{124 a q}{13}=19 a+\frac{a q^{2}}{13} \Leftrightarrow q^{2}-248 q+247=0$, from which $q=1$ or $q=247$. A decreasing geometric progression can only occur when $... | 247 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,147 |
6.1. (12 points) Yura has unusual clocks with several minute hands moving in different directions. Yura calculated that the minute hands coincided in pairs exactly 54 times in one hour. What is the maximum number of minute hands that can be on Yura's clocks?
# | # Answer: 28.
Solution. Let two arrows coincide, then after 30 seconds they will coincide again. Therefore, the arrows in each such pair will coincide exactly 2 times per minute. Thus, if $n$ arrows move in one direction, and $m$ arrows move in the other, then $2 m n=54, m n=27$. Therefore, $n$ can be $1,3,9$ or 27. T... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,148 |
7.1. (12 points) Two people spend time playing a game: they take turns naming prime numbers not exceeding 100 such that the last digit of the number named by one player is equal to the first digit of the number named by the next player (except for the first prime number named in the game). Repeating numbers that have a... | Solution. Let's describe a winning strategy for the first player. First, the first player names a prime number ending in 9 and different from 79 (for example, 19). Since among the numbers 90, ..., 99, the only prime number is 97, the second player must name this number on their next turn. Then, on the third move, the f... | 3 | Number Theory | proof | Yes | Yes | olympiads | false | 6,149 |
8.1. (12 points) In how many ways can eight of the nine digits $1,2,3,4,5,6$, 7,8 and 9 be placed in a $4 \times 2$ table (4 rows, 2 columns) so that the sum of the digits in each row, starting from the second, is 1 more than in the previous one? | # Answer: 64.
Solution. The sum of all nine numbers is 45. Let $x$ be the sum of the two numbers in the first row, and let $a$ be the one number out of the nine that we do not place in the figure. Then $x+x+1+x+2+x+3=45-a$, from which $4 x+a=39$. Since $a$ is an integer from 1 to 9, we get 2 possible cases: either $x=... | 64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,150 |
9.1. (12 points) Two circles touch each other externally at point $K$. On their common internal tangent, point $P$ is marked such that $K P=14$. Through point $P$, two secants are drawn to the circles, with one of them intercepting a chord $A B=45$ on the first circle, and the other intercepting a chord $C D=21$ on the... | Answer: 1.75.
Solution. By the tangent-secant theorem, we find $A P=4, C P=7$. Moreover, $P A \cdot P B=$ $P K^{2}=P C \cdot P D$, so triangles $B P C$ and $D P A$ are similar by two proportional sides and the angle between them. Therefore, $B C: A D=C P: A P=\frac{7}{4}$. | 1.75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,151 |
10.1. (12 points) For different natural numbers $k, l, m, n$, it is known that there exist such three natural numbers $a, b, c$ that each of the numbers $k, l, m, n$ is a root of either the equation $a x^{2}-b x+c=0$, or the equation $c x^{2}-16 b x+256 a=0$. Find $k^{2}+l^{2}+m^{2}+n^{2}$. | Answer: 325.
Solution. If $k, l$ are the roots of the first equation, then the roots of the second equation are the numbers $m=\frac{16}{k}$, $n=\frac{16}{l}$. Therefore, the numbers $k, l, m, n$ are divisors of the number 16. The divisors of 16 are the numbers 1, 2, 4, 8, and 16, but the number 4 does not fit, as by ... | 325 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,152 |
Task 1. Every time my brother tells the truth, our grandmother sneezes. One day, my brother said he got a "5" in math, but grandmother didn't sneeze. Then, slightly doubting his first words, he said he got a "4," and grandmother sneezed. Encouraged by grandmother's sneeze, he confirmed that he definitely got no less th... | Answer: $« 2 »$.
Solution. If the grandmother did not sneeze, then the brother definitely lied, so he did not get a "5" and even more than that, less than 3. If the grandmother did sneeze, then it is not certain that he told the truth at that moment, as the condition does not prohibit the grandmother from sneezing whe... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,155 |
Problem 2. Baba Yaga must arrive at Bald Mountain exactly at midnight. She calculated that if she flies on a broom at a speed of 50 km/h, she will be 2 hours late, and if she flies on an electric broom at a speed of 150 km/h, she will arrive 2 hours early. To arrive at Bald Mountain exactly on time, Baba Yaga used a br... | Answer: At 20:00 at a speed of 75 km/h.
Solution. The electric broom flies three times faster than the cauldron, and at the same time, it saves 4 hours of time. This means that two-thirds of the flight time in the cauldron is 4 hours, the entire flight in the cauldron would have taken 6 hours, and the entire flight on... | At20:00atspeedof75/ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,156 |
Task 3. In one school, four friends study - all in different grades: the youngest is in the first grade, and the oldest is in the fourth. Determine the name, surname, and grade of each of them, given that:
1) Borya is not a first-grader;
2) when Vasya goes to the pool on the neighboring Southern Street, Ivanov walks h... | Answer: Dima Ivanov is in the first grade, Misha Krylov is in the second, Borya Petrov is in the third, and Vasya Orlov is in the fourth.
Solution. First, let's match the boys' names with their grades. From conditions 1, 3, and 6, it follows that only Dima can be a first-grader. Then, from condition 3, Misha is in the... | DimaIvanovisinthefirstgrade,MishaKrylovisinthe,BoryaPetrovisinthethird,VasyaOrlovisinthefourth | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,157 |
Problem 4. Vanya thought of a two-digit number, then swapped its digits and multiplied the resulting number by itself. The result turned out to be four times larger than the number he thought of. What number did Vanya think of? | Answer: 81.
Solution. Let the intended number be $\overline{m n}=10 m+n$. Then $4 \overline{m n}=\overline{n m}^{2}$. Therefore, $\overline{n m}^{2}$ is divisible by 4, and $\overline{n m}$ is divisible by 2, so the digit $m$ is even (and not zero). Moreover, $\overline{m n}=\overline{n m}^{2}: 4=(\overline{n m}: 2)^{... | 81 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,158 |
Problem 5. The Martian traffic light consists of six identical bulbs arranged in two horizontal rows (one above the other) with three bulbs in each. A rover driver in the fog can distinguish the number and relative position of the lit bulbs on the traffic light (for example, if two bulbs are lit, whether they are in th... | Answer: 44.
Solution. If two traffic light signals differ only by the shift of the lit bulbs, then the driver cannot distinguish them (and vice versa). Therefore, any signal can either be transformed by a shift to the left and/or up into an indistinguishable signal, which has at least one bulb lit in the top horizonta... | 44 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,159 |
1.1. (2 points) The average age of employees in a company consisting of 13 people is 35 years. After a new employee is hired, the average age of the employees becomes 34 years. Find the age of the new employee. | Answer: 21.
Solution. The sum of the employees' ages before the new hire was $13 \cdot 35$, and after the new employee was hired, the sum of the ages became $14 \cdot 34$. Therefore, the age of the new employee is $14 \cdot 34 - 13 \cdot 35 = 35 - 14 = 21$. | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,160 |
2.1. (2 points) On the side $B C$ of triangle $A B C$, a point $D$ is chosen such that $\angle B A D=50^{\circ}$, $\angle C A D=20^{\circ}$, and $A D=B D$. Find $\cos \angle C$. | Answer: $\frac{\sqrt{3}}{2} \approx 0.87$.
Solution. Triangle $A B D$ is isosceles, $\angle A D B=\angle B A D=50^{\circ}$, then $\angle A B D=80^{\circ}$, $\angle B A C=70^{\circ}$, therefore $\angle C=180^{\circ}-70^{\circ}-80^{\circ}=30^{\circ}, \cos \angle C=\frac{\sqrt{3}}{2} \approx 0.87$. | \frac{\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,161 |
3.1. (12 points) An eraser, 3 pens, and 2 markers cost 240 rubles. Two erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers? | Answer: 520.
Solution. From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will cost $2 \cdot 240=480$ rubles. Therefore, one pen costs $480-440=40$ rubles. Then, 3 erasers, 4 pens, and 6 markers cost $680-4 \cdot 40=520$ rubles. | 520 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,163 |
4.1. (12 points) In an acute-angled triangle $A B C$, angle $A$ is $35^{\circ}$, segments $B B_{1}$ and $C C_{1}$ are altitudes, points $B_{2}$ and $C_{2}$ are the midpoints of sides $A C$ and $A B$ respectively. Lines $B_{1} C_{2}$ and $C_{1} B_{2}$ intersect at point $K$. Find the measure (in degrees) of angle $B_{1}... | # Answer: 75.
Solution. Note that angles $B$ and $C$ of triangle $ABC$ are greater than $\angle A=35^{\circ}$ (otherwise it would be an obtuse triangle), so point $C_{1}$ lies on side $AB$ b... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,164 |
5.1. (12 points) The equation $x^{2}+5 x+1=0$ has roots $x_{1}$ and $x_{2}$. Find the value of the expression
$$
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}
$$ | Answer: 220.
Solution. Since $x_{1}^{2}=-5 x_{1}-1$, then $\left(1+x_{1}\right)^{2}=1+2 x_{1}-5 x_{1}-1=-3 x_{1}$. Therefore,
$$
\begin{gathered}
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}=6\left(\frac{-5 x_{1}-1}{-3 x_{2}}+\frac{-5 x_{2}-1}{-3 x_{1}}\right)=\frac{... | 220 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,165 |
6.1. (12 points) From point $A$ to point $B$, a bus and a cyclist departed simultaneously at 13:00. After arriving at point $B$, the bus, without stopping, headed back and met the cyclist at point $C$ at 13:10. Upon returning to point $A$, the bus again, without stopping, headed to point $B$ and caught up with the cycl... | Answer: 40.
Solution. Let $v_{1}$ and $v_{2}$ be the speeds (in km/h) of the cyclist and the bus, respectively. By the time of the first meeting, they have collectively traveled $\frac{1}{6} v_{1}+\frac{1}{6} v_{2}=8$ km. Until the next meeting, the time that has passed is equal to $\frac{s_{0}}{v_{1}}=\frac{2 \cdot \... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,166 |
7.1. (12 points) The numbers $a$ and $b$ are such that the polynomial $x^{4}+x^{3}+2 x^{2}+a x+b$ is the square of some other polynomial. Find $b$. | Answer: $\frac{49}{64} \approx 0.77$.
Solution. The polynomial whose square is the given one is a quadratic trinomial with a coefficient of 1 or -1 for $x^{2}$. We can assume that the leading coefficient is 1 (otherwise, factor out -1 with a change in the signs of the other coefficients). Let $\left(x^{2}+A x+B\right)... | \frac{49}{64} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,167 |
8.1. (12 points) In triangle $ABC$, the bisector $BL$ is drawn. Find the area of the triangle if it is known that $AL=2, BL=3\sqrt{10}$, and $CL=3$. | Answer: $\frac{15 \sqrt{15}}{4} \approx 14.52$.
Solution. By the property of the angle bisector of a triangle $A B: B C=A L: L C=2: 3$. Therefore, if $A B=2 x$, then $B C=3 x$. Then, since $B L^{2}=A B \cdot B C-A L \cdot L C$, we get $90=6 x^{2}-6$, from which $x=4$. Thus, triangle $A B C$ has sides 8, 12, and 5. Fro... | \frac{15\sqrt{15}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,171 |
10.1. (12 points) Find the minimum value of the function $f(x)=x^{2}+3 x+\frac{6}{x}+\frac{4}{x^{2}}-1$ on the ray $x>0$. | Answer: $3+6 \sqrt{2} \approx 11.49$.
Solution. For $x>0$ we have $t=x+\frac{2}{x} \geqslant 2 \sqrt{2}$ (equality is achieved at $x=\sqrt{2}$), and in this case $f(x)=\left(x+\frac{2}{x}\right)^{2}-3\left(x+\frac{2}{x}\right)-5$. Therefore, we need to find the minimum value of the function $g(t)=t^{2}-3 t-5$ for $t \... | 3+6\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,175 |
10.4. Find the minimum value of the function $f(x)=x^{2}+2 x+\frac{6}{x}+\frac{9}{x^{2}}+4$ on the ray $x>0$. | Answer: $10+4 \sqrt{3} \approx 16.93$. | 10+4\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,177 |
1.1. (2 points) In a nine-story building, there are 4 apartments on each floor. How many entrances are there in this building if there are a total of 180 apartments? | Answer: 5.
Solution. The number of entrances is $180:(4 \cdot 9)=5$. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,178 |
2.1. (14 points) Mitya is 11 years older than Shura. When Mitya was as old as Shura is now, he was twice as old as she was. How old is Mitya? | Answer: 33.
Solution. 11 years ago, Shura was half as old as she is now. So, she is 22 years old, and Mitya is 33. | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,179 |
3.1. (14 points) Represent the number $\frac{201920192019}{191719171917}$ as an irreducible fraction. In the answer, write down the denominator of the resulting fraction. | Answer: 639.
Solution. We have
$$
\frac{201920192019}{191719171917}=\frac{2019 \cdot 100010001}{1917 \cdot 100010001}=\frac{2019}{1917}=\frac{3 \cdot 673}{3 \cdot 639}=\frac{673}{639}
$$
Since $639=3^{2} \cdot 71$ and 673 is not divisible by 3 and 71, the resulting fraction is irreducible. | 639 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,180 |
4.1. (14 points) How many six-digit numbers exist whose sum of digits equals 51? | Answer: 56.
Solution. Since the maximum sum of the digits of a six-digit number is $6 \cdot 9=54$, the sought numbers are $699999, 789999, 888999$, as well as all numbers obtained from them by rearranging the digits. The first of these numbers can form 6 six-digit numbers (the digit 6 can be in any of the 6 positions)... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,181 |
5.1. (14 points) The old man was pulling the turnip, and one by one, the old woman, the granddaughter, the dog, and the cat joined him. They pulled and pulled, but couldn't pull out the turnip! The cat called the mouse. They pulled and pulled, and finally pulled out the turnip! It is known that each subsequent particip... | Answer: 2.
Solution. The old woman, the granddaughter, Zhuchka, and the mouse pull with a combined force of
$$
\frac{3}{4}+\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{4}\right)^{3}+\left(\frac{3}{4}\right)^{5}=\frac{2019}{1024}=1 \frac{995}{1024}
$$
of the force of the old man. Therefore, two men need to be called. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,182 |
6.1. (14 points) Find the smallest natural number whose representation contains 5 zeros and 7 ones, and the sum of the digits in the even positions is equal to the sum of the digits in the odd positions. | Answer: 1000001111131.
Solution. The number 7 is not divisible by 2, so the number cannot have exactly $5+7=12$ digits. Consider the case where the number has 13 digits. If the "new" digit is zero, we again get a contradiction. Therefore, the "new" digit must be at least 1. Thus, to find the smallest number, we choose... | 1000001111131 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,183 |
7.1. (14 points) Misha drew a triangle with a perimeter of 11 and cut it into parts with three straight cuts parallel to the sides, as shown in the figure. The perimeters of the three shaded figures (trapezoids) turned out to be 5, 7, and 9. Find the perimeter of the small triangle that resulted from the cutting.
# | # Answer: 10.
Solution. Since opposite sides of parallelograms are equal, the lateral sides of each resulting trapezoid are equal to the corresponding segments on the sides of the original tr... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,184 |
8.1. (14 points) Two people spend time playing a game: they take turns naming prime numbers not exceeding 100 such that the last digit of the number named by one player is equal to the first digit of the number named by the next player (except for the first prime number named in the game). Repeating numbers that have a... | # Answer: 3.
Solution. We will describe a winning strategy for the first player. First, the first player names a prime number ending in 9 and different from 79 (for example, 19). Since among the numbers $90, \ldots, 99$ the only prime is 97, the second player must name this number on their next move. Then, on the thir... | 3 | Number Theory | proof | Yes | Yes | olympiads | false | 6,185 |
# Task 1.
Mom told Dima that he should eat 13 spoons of porridge. Dima told his friend that he had eaten 26 spoons of porridge. Each child, telling about Dima's feat, increased the amount of porridge Dima had eaten by 2 or 3 times. As a result, one of the children told Dima's mom about 33696 spoons of porridge. How ma... | Answer: 9
Solution. $33696=2^{5} \cdot 3^{4} \cdot 13$. From this, it follows that the story of the feat was retold 5 times with the amount of porridge doubling and 4 times with it tripling, totaling 9 times.
## B-2
Dad persuaded Natasha to eat 11 spoons of porridge. Natasha told her friend that she had eaten 22 spo... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,186 |
# Task 2.
B-1.
At the time when a lion cub, who was 5 minutes away, set off for the watering hole, the second cub, having already quenched its thirst, started heading back along the same road at 1.5 times the speed of the first. At the same time, a tortoise, which was half an hour away, set off for the watering hole ... | Answer: 28
Solution. Let's take the entire path of the turtle as 1 and let $x$ be the speed of the 1st lion cub. Then the speed of the 2nd lion cub is $1.5x$, and the speed of the turtle is $1/30$. The entire path to the watering hole for the 1st lion cub is $5x$. Therefore, the meeting with the 2nd lion cub occurred ... | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,187 |
# Problem 3.
A square with a side of 75 mm was cut by two parallel cuts into three rectangles. It turned out that the perimeter of one of these rectangles is half the sum of the perimeters of the other two. What is this perimeter? Give your answer in centimeters. | # Answer: 20
Solution. For three rectangles, one of the sides is the same and equals 75 mm. The sum of the lengths of two such sides of one rectangle is half the sum of the lengths of all such sides of the other two. Therefore, the other side of the first rectangle is also half the sum of the other sides of the two ot... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,188 |
# Task 4.
Find the smallest 12-digit natural number that is divisible by 36 and contains all 10 digits in its decimal representation. | Answer: 100023457896
Solution. The number is divisible by 4 and 9. Since the sum of ten digits is 45 (divisible by 9), two more digits should be added to these ten digits, the sum of which is 0, 9, or 18. Since we need the smallest number, we will add two digits 0 and place the number 10002345 at the beginning of the ... | 100023457896 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,189 |
# Problem 5.
B-1
Chicks hatch on the night from Sunday to Monday. For two weeks, the chick sits with its beak open, the third week it silently grows feathers, and on the fourth week, it flies away from the nest. Last week, 20 chicks were sitting with their beaks open, and 14 were growing feathers, while this week, 15... | Answer: 225
Solution. In fact, the chicks should be divided into three categories: one-week-old, two-week-old, and three-week-old. With each new week, each chick moves to the next category. So, if 11 are feathering this week, then last week there were 11 two-week-olds, and, accordingly, 9 one-week-olds ( $9+11=20$ ). ... | 225 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,190 |
# Problem 6.
In the alphabet of the inhabitants of the magical planet ABV2020, there are only three letters: A, B, and V, from which all words are formed. In any word, two identical letters cannot be adjacent, and each of the three letters must be present in any word. For example, the words ABV, VABAVAB, and BVBVAB ar... | # Answer: 1572858
Solution. The first letter can be any of the three, and for each subsequent letter, there are two options. This results in $3 \cdot 2^{19}$ words. However, we need to subtract from this the words that are made up of only two letters, not three. There are 6 such words. Therefore, we get $3 \cdot 2^{19... | 1572858 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,191 |
# Task 7.
Around a round table, 1001 people are sitting, each of whom is either a knight (always tells the truth) or a liar (always lies). It turned out that next to each knight sits exactly one liar, and next to each liar there is a knight. What is the minimum number of knights that can sit at the table? | Answer: 502
Solution. From the condition, it follows that a knight cannot sit between two knights or two liars, and a liar cannot sit between two liars. Thus, when moving around the table, knights will be encountered in pairs, while liars will be encountered singly or in pairs. From this, it follows that each liar can... | 502 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,192 |
1. On the Island of Knights and Liars, there live knights who always tell the truth and liars who always lie. One day, five residents of this island were asked one by one how many knights were among them.
- One, - answered the first.
- Two, - answered the second.
- Three, - answered the third.
- Don't believe them, th... | Solution. Among the first three, there is no more than one knight, since they all give different answers. If we consider the 4th and 5th, one of them is a knight and the other is a liar. Therefore, there is one or two knights in total. In any case, one of the first three told the truth. This means there are two knights... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,193 |
3. Let's call a natural number $n$ squareable if the numbers from 1 to $n$ can be arranged in such an order that each member of the sequence, when added to its position, results in a perfect square. For example, the number 5 is squareable, as the numbers can be arranged as: 32154, in which $3+1=2+2=1+3=4$ and $5+4=4+5=... | # Answer: 9 and 15.
Solution. The number 7 cannot be squareable, since both numbers 1 and 6 must be in the third position, which is impossible.
The number 9 is squareable, as the numbers from 1 to 9 can be arranged in the following order: $8,2,6$, $5,4,3,9,1,7$, thus satisfying the required condition.
The number 11 ... | 915 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,194 |
4. The numbers from 1 to 8 are arranged at the vertices of a cube such that the sum of the numbers in any three vertices lying on the same face is at least 10. What is the smallest possible sum of the numbers at the vertices of one face? | Answer: 16.
Solution. Each face has a vertex where a number not less than 6 is placed. Indeed, otherwise, one of the triples of the remaining largest numbers $2,3,4,5$ would give a sum less than 10 (specifically, the triple $2,3,4$ with a sum of 9).
Consider a face containing the vertex where the number 6 is placed. ... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,195 |
5. On graph paper (the side of a cell is 1 cm) a rectangle is drawn, the sides of which lie on the grid lines, with one side being 7 cm shorter than the other. It turned out that it can be cut into several pieces along the grid lines and form a square from them. What can the side of this square be? Find all possible va... | Answer: 6 cm.
Solution. Let the larger side of the rectangle be $k$ cm, where $k>7$, and the side of the resulting square be $n$ cm. Then, from the equality of the areas, we get $k(k-7)=n^{2}$. Since $n<k$, let $n=k-m$, where $m \geqslant 1$. Then $k^{2}-7 k=(k-m)^{2}, k^{2}-7 k=k^{2}-2 m k+m^{2}$, from which $m(2 k-m... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,196 |
6. All natural numbers, the sum of the digits of each of which is equal to 5, were arranged in ascending order. What number is in the 125th place | Answer: 41000.
Solution. Let's calculate the number of $n$-digit numbers, the sum of the digits of each of which is equal to 5, for each natural $n$. Subtract 1 from the leading digit, we get a number (which can now start with zero), the sum of the digits of which is equal to 4. Represent the digits of this number as ... | 41000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,197 |
7. In "Dragon Poker," the deck has four suits. An Ace brings 1 point, a Jack -2 points, a Two $-2^{2}$, a Three $-2^{3}$, ..., a Ten $-2^{10}=1024$ points. Kings and Queens are absent. You can choose any number of cards from the deck. In how many ways can you score 2018 points? | Answer: $C_{2021}^{3}=1373734330$.
Solution. In any suit, you can score any number of points from 0 to 2047, and this can be done in a unique way. This can be done as follows: write down this number in binary and select the cards corresponding to the binary positions containing 1 (i.e., if there is a 1 in the $k$-th p... | 1373734330 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,198 |
# Problem 1.
It is known that the numbers $\frac{x}{2}, 2 x-3, \frac{18}{x}+1$, taken in the given order, form a geometric progression. Find the common ratio of this progression. Round your answer to two decimal places. | Answer: $\frac{52}{25}=2.08$
Solution. By the characteristic property of a geometric progression $a_{n} \cdot a_{n+2}=a_{n+1}^{2}$. We get the equation: $\frac{x}{2} \cdot\left(\frac{18}{x}+1\right)=(2 x-3)^{2}$. Hence $8 x^{2}-25 x=0$. At $x=0$ the domain is violated, and at $x=\frac{25}{8}$ we get a geometric progre... | 2.08 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,199 |
# Task 2.
While a lion cub, located 6 minutes away, was heading to the watering hole, the second one, having already quenched his thirst, set off along the same path back at 1.5 times the speed of the first. At the same time, a turtle, located 32 minutes away, set off along the same path to the watering hole. After so... | Answer: 28.8
Solution. Let $x$ be the speed of the turtle, and $y$ be the speed of the first lion cub. Then the speed of the second lion cub is $1.5 y$. The entire path to the watering hole for the first lion cub is $6 y$, and for the turtle, it is $32 x$. Therefore, the initial distance between them was $6 y - 32 x$,... | 28.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,200 |
# Task 3.
Masha tightly packed 165 identical balls into the shape of a regular triangular pyramid. How many balls lie at the base? | Answer: 45
Solution. We will solve the problem in general for $\frac{1}{6} n(n+1)(n+2)$ balls. At the base of a triangular pyramid of height $n$ rows lies the $n$-th triangular number of balls $1+2+3+\ldots+n=\frac{1}{2} n(n+1)$. Then the total number of balls in the pyramid is $\sum_{k=1}^{n} \frac{k(k+1)}{2}=\frac{1... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,201 |
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